Question

1. sketch the algebra tiles for each trinomial arranged as a rectangle to show how they can be factored. (a) ( x ^ { 2 } + 7 x + 10 )

Explanation:

Step1: Factor the trinomial

We need to find two numbers that multiply to \(10\) (the constant term) and add up to \(7\) (the coefficient of the middle term). The numbers \(2\) and \(5\) satisfy these conditions since \(2\times5 = 10\) and \(2 + 5=7\). So, \(x^{2}+7x + 10=(x + 2)(x+5)\).

Step2: Sketch the algebra - tiles

• Represent \(x^{2}\) with one large square tile.
• Represent \(7x\) with seven rectangular tiles (where the length is \(x\) and the width is \(1\)).
• Represent \(10\) with ten small square tiles (with side - length \(1\)).

Arrange them into a rectangle. The length of the rectangle will be \((x + 5)\) and the width will be \((x + 2)\). One side of the rectangle has \(x\) (from the \(x^{2}\) tile) and \(5\) (from the \(5\) unit - square tiles), and the other side has \(x\) (from the \(x^{2}\) tile) and \(2\) (from the \(2\) unit - square tiles).

Answer:

The trinomial \(x^{2}+7x + 10\) can be factored as \((x + 2)(x + 5)\). To sketch the algebra - tiles: use 1 \(x^{2}\) tile, 7 \(x\) tiles, and 10 unit - square tiles. Arrange them into a rectangle with length \((x + 5)\) and width \((x + 2)\).